The tangent to the curve 〖y=e〗^2x at the point (0, 1) meets x-axis at:
(A)(0, 1) (B) − 1/2, 0
(C) (2, 0) (D) (0, 2)



Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
NCERT Exemplar - MCQs
Question 2 Important
Question 3 Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 9 Important
Question 10
Question 11 Important
Question 12 Important You are here
Question 13
Question 14
Question 15 Important
Question 16 Important
Question 17 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams You are here
Question 13 Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Question 12 The tangent to the curve 〖𝑦=𝑒〗^2𝑥 at the point (0, 1) meets x-axis at: (0, 1) (B) − 1/2, 0 (C) (2, 0) (D) (0, 2) Given 〖𝑦=𝑒〗^2𝑥 Finding slope of tangent at (0, 1) 〖𝑦=𝑒〗^2𝑥 Differentiating w.r.t. 𝑥 𝒅𝒚/𝒅𝒙 = 2𝒆^𝟐𝒙 At point (0, 1) Putting 𝑥=0, 𝑦=0 in 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 〖2𝑒〗^0 𝑑𝑦/𝑑𝑥 = 2 × 1 𝒅𝒚/𝒅𝒙=𝟐 Finding equation of tangent Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) ∴ Equation of tangent at point (0, 1) & having slope 2 is (𝒚−𝟏) = 2 (𝒙−𝟎) (𝑦−1) = 2𝑥 𝒚=𝟐𝒙+𝟏 Now, we need to find the point where tangent meets the x-axis Since point is on the x-axis, it’s y-coordinate will be 0 ∴ Point = (𝒙,𝟎) Putting 𝑦 = 0 in (1) 0 = 2𝒙 + 1 −1 = 2𝑥 (−1)/2=𝑥 𝒙 = (−𝟏)/𝟐 Thus, required point is ((−𝟏)/𝟐, 𝟎) So, the Correct answer is (B)